I'm pretty sure my question is not well put so I will try to explain it little bit more. Let's say I have a Attiny85 without any firmware. What kind of firmware do I need to flash Attiny with, in order for Windows OS to recognize it as serial device connected via USB and assign it COM port for communication?

I don't want to install any additional drivers in OS, I would like Windows to use already preinstalled CH341S64 winchiphead drivers for USB to serial conversion.

Also I won't use any kind of bootloader on the Attiny itself for prototyping.

  • 3
    \$\begingroup\$ It needs to implement the USB communications device class (CDC). Or perhaps your board has some USB-to-Serial converter such as one of the FTDI family, which is implementing the mentioned CDC. \$\endgroup\$ – Eugene Sh. May 22 '18 at 18:28
  • 1
    \$\begingroup\$ Arduino uses a seperate USB->UART Chip. The usb is not directly connected to atmega328p \$\endgroup\$ – Daniel Johnson May 22 '18 at 18:46
  • 1
    \$\begingroup\$ Arduino has a USB-Serial chip that the PC recognizes. If you want to connect only with the ATTiny85 you need some sort of software loaded. Start here: arduino.stackexchange.com/questions/20240/… \$\endgroup\$ – Jack Creasey May 22 '18 at 18:48
  • 1
    \$\begingroup\$ @SirDrinksCoffeeALot an alternative part in the same family the ATmega16u2 offers on board USB and DFU support. This part is used as the programming/serial-port device for the 32-bit Arduino Zero \$\endgroup\$ – crasic May 22 '18 at 18:59
  • 1
    \$\begingroup\$ For the record, neither FTDI nor (AFAIK) WinChipHead nor Prolific try to mimick the CDC ACM class. They each have their respective proprietary USB ID's and need a proprietary driver. But, many other USB/serial gadgets choose to behave like a CDC device. For a broader explanation, see e.g. rfc1149.net/blog/2013/03/05/… \$\endgroup\$ – frr May 22 '18 at 19:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.